Problem: This is the easy version of the problem. The difference between the versions is that in the easy version all prices a_i are different. You can make hacks if and only if you solved both versions of the problem.

Today is Sage's birthday, and she will go shopping to buy ice spheres. All n ice spheres are placed in a row and they are numbered from 1 to n from left to right. Each ice sphere has a positive integer price. In this version all prices are different.

An ice sphere is cheap if it costs strictly less than two neighboring ice spheres: the nearest to the left and the nearest to the right. The leftmost and the rightmost ice spheres are not cheap. Sage will choose all cheap ice spheres and then buy only them.

You can visit the shop before Sage and reorder the ice spheres as you wish. Find out the maximum number of ice spheres that Sage can buy, and show how the ice spheres should be reordered.

Input

The first line contains a single integer n (1 ≤ n ≤ 10^5) — the number of ice spheres in the shop.

The second line contains n different integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9) — the prices of ice spheres.

Output

In the first line print the maximum number of ice spheres that Sage can buy.

In the second line print the prices of ice spheres in the optimal order. If there are several correct answers, you can print any of them.

Example

Input


5
1 2 3 4 5


Output


2
3 1 4 2 5 

Note

In the example it's not possible to place ice spheres in any order so that Sage would buy 3 of them. If the ice spheres are placed like this (3, 1, 4, 2, 5), then Sage will buy two spheres: one for 1 and one for 2, because they are cheap.

Solution (in JAVA):

Solution:
import java.io.*;
import java.util.*;

public class CFA {
  static final PrintWriter out = new PrintWriter(System.out);
  // static final Reader in = new Reader();
  static final Scanner in = new Scanner(System.in);

  static class Reader {
    BufferedReader br;
    StringTokenizer st;

    public Reader() {
      br = new BufferedReader(new InputStreamReader(System.in));
    }

    String next() {
      while (st == null || !st.hasMoreElements()) {
        try {
          st = new StringTokenizer(br.readLine());
        } catch (final IOException e) {
          e.printStackTrace();
        }
      }
      return st.nextToken();
    }

    int nextInt() {
      return Integer.parseInt(next());
    }

    long nextLong() {
      return Long.parseLong(next());
    }

    double nextDouble() {
      return Double.parseDouble(next());
    }

    String nextLine() {
      String str = "";
      try {
        str = br.readLine();
      } catch (final IOException e) {
        e.printStackTrace();
      }
      return str;
    }
  }

  static long binpow(final int a, final int b) {
    if (b == 0) return 1;
    final long res = binpow(a, b / 2);
    if (b % 2 == 1) return res * res * a; else return res * res;
  }

  static int gcd(final int a, final int b) {
    if (b == 0) return a;
    return gcd(b, a % b);
  }

  static void println(String s) {
    System.out.println(s);
  }

  static void print(String s) {
    System.out.print(s);
  }

  static int n, k, a[], b[];
  static Map<Integer, Integer> map;
  static char[] w;
  static int x;

  public static void main(final String[] args) throws Exception {
    n = in.nextInt();
    a = new int[n];
    b = new int[n];

    for (int i = 0; i < n; i++) a[i] = in.nextInt();

    Arrays.sort(a);

    int j = n - 1;
    int k = 0;
    for (int i = 0; i < n; i += 2, j--) {
      b[i] = a[j];
    }

    if (n % 2 == 0) {
      
      b[n-1] = a[j];
      j--;

      for(int i = n-3;i>=0;i-=2,k++){
        b[i] = a[k];
      }
    } else{
      
      for(int i = n-2;i>=0;i-=2,k++){
        b[i] = a[k];
      }
    }   

    int cnt = 0;
    for(int i = 1;i < n-1;i++){
      if(b[i] < b[i-1] && b[i] < b[i+1]) 
        cnt++;
    }

    out.println(cnt);
    for(int i = 0;i < n;i++){
      out.print(b[i]+" ");
    }

    out.println();
    out.close();
  }
}